Answer
These reactants are capable of producing $2.33$ g of $BaSO_4$;
Work Step by Step
1. Write and balance the reaction between $BaCl_2$ and $Fe_2(SO_4)_3$:
$BaCl_2(aq) + Fe_2(SO_4)_3(aq) -- \gt BaSO_4(s) +FeCl_3(aq)$
- Balance it:
$3BaCl_2(aq) + Fe_2(SO_4)_3(aq) -- \gt 3BaSO_4(s) +2FeCl_3(aq)$
2. Find the number of moles of $BaSO_4$ that each compound can produce.
$100.0mL \times \frac{1L}{1000mL} \times \frac{0.100mol(BaCl_2)}{1L} \times \frac{3mol(BaSO_4)}{3mol(BaCl_2)} = 0.0100mol(BaSO_4)$
$100.0mL \times \frac{1L}{1000mL} \times \frac{0.100mol(Fe_2(SO_4)_3)}{1L} \times \frac{3mol(BaSO_4)}{1mol(Fe_2(SO_4)_3)} =0.0300mol(BaSO_4)$
- $BaCl_2$ produces less barium sulfate, so, it is the limiting reactant.
- Use the number of moles produced by $BaCl_2$ to calculate the mass:
Molar mass: 137.3* 1 + 32.07* 1 + 16.00* 4 = 233.4g/mol $(BaSO_4)$
$0.0100 mol(BaSO_4) \times \frac{233.4g (BaSO_4)}{1mol(BaSO_4)} = 2.33g(BaSO_4)$
